200 `cm^3` of an aqueous solution of a protein contains 1.26g of the protein . The osmotic pressure of such a solution at 300K is found to be `2.7 xx 10^-3` bar. Calculate the molar mass of the protein (R=0.083 L bar `mol^-1 K^-1`)

Explanation

Mass of solute = 1.26 g
Volume of solution = 200 `"cm"^(3)`
Osmotic pressure = `2.7xx10^(-3)"bar"`
Molar mass of solute = ?
`pi=CRT,2.7xx10^(-3)=1.23/(M)xx1000/(200)xx0.083xx300`
`M=(1.26xx1000xx0.083xx300)/(2.7xx10^(-3)xx200)=58.1xx10^(3)=58100"g mol"^(-1)`